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I'm trying my hand on these types of expressions. How to solve $x$ in $(x+1)^4+(x-1)^4=16$? Please write any idea you have, and try to keep it simple. Thanks.

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We have our equation to solve for: $$(x+1)^4+(x-1)^4=16$$ First, expand the terms in the left hand side. $$x^4+4x^3+6x^2+4x+1+x^4-4x^3+6x^2-4x+1=16$$ Simplify the left hand side. $$2x^4+12x^2+2=16$$ Factor $2$ out from both sides. $$x^4+6x^2+1=8$$ Move $8$ to the left hand side. $$x^4+6x^2-7=0$$ Let $x^2=a$. Now our equation becomes: $$a^2+6a-7=0$$ Factor it. $$(a-1)(a+7)=0$$ Reverse the substitution. $$(x^2-1)(x^2+7)=0$$ Factor some more (yes even the sum of squares) $$(x+1)(x-1)(x+\sqrt7i)(x-\sqrt7i)=0$$ The four roots are $x=\pm 1$ and $x=\pm \sqrt{7}i$. But if you do not want complex roots, then the two real roots are $x=1$, $x=-1$.

Hope I helped!

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  • $\begingroup$ The left hand side should simplify to $2x^4+12x^2+2$ in your first step. $\endgroup$ – Sujaan Kunalan Apr 5 '14 at 17:26
  • $\begingroup$ Oh woops thanks for the correction $\endgroup$ – TrueDefault Apr 5 '14 at 17:27
  • $\begingroup$ No problem. Looks good now. (+1) $\endgroup$ – Sujaan Kunalan Apr 5 '14 at 17:36
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HINT:

Just expand in Binomial Expansion to find a Quadratic Equation in $x^2$

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Using the expansion $$\left(a+b\right)^n = \sum_{r=0}^n\binom{n}{r}a^rb^{n-r}$$

$$\require{cancel}{x^4+\cancel{4x^3}+6x^2+\cancel{4x}+1+x^4+-\cancel{4x^3}+6x^2-\cancel{4x}+1 = 16}$$

$$x^4+6x^2+1=8$$

$$t^2+6t-7=0\quad\text{with}\quad x^2=t$$

$$t=1,-7$$

Since $x^2=-7$ doesn't have solutions in reals, $x^2=1\implies x=\pm1$

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Let denote by $P(X)$ the following polynomial:

$$P(X)=(X+1)^4+(X-1)^4-16.\quad\quad(1)$$

We immediately see that $X=1$ and $X=-1$ are roots of $P(X)$. Therefore, $P(X)$ can be written as:

$$P(X)=(X-1)(X+1)Q(X), \quad\quad\;\;\;\;\;\;(2)$$ where $Q(X)$ is a polynomial with degree $2$. Hence, $Q(X)=aX^2+bX+c$ for some $a$, $b$, and $c$.

Now, if we develop $P(X)$ for both $(1)$ and $(2)$ we get:

First, with $(1)$: $$P(X)=2X^4+12X^2-14.\quad\quad(3)$$ Second, with $(2)$: $$P(X)=(X-1)(X+1)(aX^2+bX+c)=aX^4+bX^3+(c-a)X^2-bX-c.\quad(4)$$ Hence, $(3)=(4)$

$$2X^4+12X^2-14=aX^4+bX^3+(c-a)X^2-bX-c.$$

Then, $$ \begin{array}{lr} a=2\\ b=0\\ c=14 \end{array}$$

And finally,

$$P(X)=(X+1)(X-1)(2X^2+14)=2(X+1)(X-1)(X^2+7).$$

Now it is clear how to solve $P(X)=0$ (how to find the two other roots):

$$S_{\mathbb{R}}=\{-1, +1\}.$$ and $$S_{\mathbb{C}}=\{-1, +1, +i\sqrt{7}, -i\sqrt{7}\}.$$

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Looking at $(x+1)^4 + (x-1)^4=16$, I recall that $16=2^4$. So I look at the equation trying to see if I could get a $2^4$ in there. This sugguests that we guess $x=1$ to make the left term a power of $2$.

Plugging in $x=1$,

$$ (1+1)^4 + (1-1)^4 = 2^4 + 0^4 = 16 $$

So $x=1$ is a solution.

This problem has a symmetry property. Notice that sending $x\rightarrow -x$ doesn't change the expression on the left.

$$ (x+1)^4 + (x-1)^4 \rightarrow (-x+1)^4 + (-x-1)^4 = (x-1)^4+(x+1)^4$$

So we know that $x=-1$ will also be a solution.

Some thought should convince you the only possible solutions are between $-1$ and $1$ since any large values of $x$ will immediately make the expression on the left bigger than $16$.

Looking at the derivative of the expression on the left,

$$ y' = 4(x+1)^3 + 4(x-1)^3, $$

we see that the derivative is $0$ at $x=0$ and monotonically increasing for $x > 0$ which means that there are no other solutions since our graph can only intersect the horizontal line $y=16$ once.

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$(x+1)^4+(x-1)^4=16\\ \implies (x^4+4x^3+6x^2+4x+1)+(x^4-4x^3+6x^2-4x+1)-16=0\\ \implies (2x^4+12x^2+2)-16\\ \implies 2x^4+12x^2-14=0\\ \implies 2(x^4+6x^2-7)=0\\ \implies 2((x^2)^2+6(x^2)^1-7(x^2)^0)=0$

Now let $x^2=a$:

$2((a)^2+6(a)^1-7(a)^0)=0\\ \implies 2(a^2+6a-7)=0\\ \implies a^2+6a-7=0\\ \implies (a+7)(a-1)=0\\ \implies a=-7 \text{ or } a=1$.

Letting $a=x^2$ again, we get

$x^2=-7\text{ or } x^2=1$.

Hence the real solutions of $x$ are $\pm1$.

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Here's the way you can deal with your exercise. I hope it will help you!

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enter image description here

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polynomial equation step by step. I hope it will help you! page 1 polynomial equation step by step page 2 polynomial equation step by step page 3 polynomial equation step by step page 4 polynomial equation step by step page 5 polynomial equation step by step page 6 polynomial equation step by step graph for this polynomial equation step by step

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  • $\begingroup$ Rather than typing this up using some other software and posting it as an image, it's much better to just post the text here (which is searchable, accessible, and takes less space). This site supports most TeX syntax through MathJax. More info on formatting can be found under "help" at the top of the page. $\endgroup$ – Alex Becker Apr 5 '14 at 22:55

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